The \(F_3\)  Appell function can be converted into a hypergeometric function. [1]} gives
\(F_3\left(\begin{array}{l}a,b,c,d \\a+b+c+d\end{array};\dfrac{1-x}{2},\dfrac{x-1}{x+1}\right)=\left(\dfrac{1+x}{2}\right)^b \,\mbox{}_{2}F_{1}\!\left(\genfrac{}{}{0.0pt}{}{a+b,b+c}{a+b+c+d};\dfrac{1-x}{2}\right)\) 
